Re: Two pair
I will take a shot at this. You have to make some assumptions that you did not give. First I will assume that you do not have a pair, and second I will assume the board does not pair.
The probability that you end up with two pair exactly hitting both of your cards is as follows.
=3*3*combin(11,3)*Power(4,3)/combin(50,5) = 4.48 %.
The second part of your question is much more difficult, you did not give enough information. Did you hit any portion of the board ie. do you have one pair. Do you want to include times that your opponent has a pair to start with. All of these will effect the probabilities of your opponents having two pair.
Assuming your opponent does not have a pair, the board does not pair, and you have not hit any portion of the board the following is the probability I got. Hopefully, someone will check my math.
One opponent>
Total hand combo's = combin(45,2) = 990
Total combo's giving a pair = combin(5,2)*9 = 90
Likely hood of one opponent pairing = 90/990 = 9.1%
This puts your opponent on a random hand at the river, obviously this is unlikely.
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